John E. McMurry
Robert C. Fay
Lecture Presentation
Chapter 14
Chemical
Equilibrium
14.1, 14.2, 14.3, 14.4,
14.7, 14.9, 14.10, 14.11,
14.12, 14.14, 14.15, 14.19,
14.23, 14.24, 14.25, 14.27,
14.31, 14.34, 14.50, 14.52,
14.54, 14.56, 14.58, 14.60.
14.68, 14.74, 14.88, 14.98,
14.100, 14.110
The Equilibrium State (kinetic chapter no reversible rxn this
chapter all reversible rxn)
Chemical Equilibrium: The state reached when the
concentrations of reactants and products remain constant
over time (dynamic equilibrium )
(dynamic equilibrium: rxn to product = rxn to reactant)
This Chapter: How far forward does the reaction go ?
NOT how fast but the ratio of products and reactants
Liquid bromine in
dynamic equilibrium
with gaseous
bromine.
The Equilibrium State
2 NO2(g)N2O4(g)
BrownColorless
The Equilibrium State 2 NO2(g)N2O4(g)
Experiment 1: forward rxn
start with only N2O4(g)
Experiment 2: backward rxn
Start with only NO2(g)
Both experiments end with exactly the same
concentration of the N2O4 and NO2
The Equilibrium State rate forward rxn = rate backward rxn
2 NO2(g)N2O4(g)
figure for rxn
Starting from N2O4
The Equilibrium Constant Kc
Why?
Equilibrium
state
Elementary Reaction Mechanism Step
rate = k1 [N2O4]
rate (back) = k2 [NO2]2
2 NO2(g)N2O4(g)
The Equilibrium Constant KcKc is the same at the same temperature no matter how much of each
reactant or product you have
cC + dDaA + bBFor a general reversible reaction:
Kc =[A]a[B]b
[C]c[D]d
Equilibrium constant expression
Equilibrium constantReactants
Products
For the following reaction: 2 NO2(g)N2O4(g)
[N2O4]
[NO2]2
= 4.64 × 10–3 (at 25 °C)Kc =
Equilibrium equation:
End 2/26W A section
HW 14.1: The Equilibrium Constant KcWrite the equilibrium constant expression for the
following reactions. (Kp = equilibrium constants with amounts given by
pressure of each component)(if gas is in solution use Kc, if gas is in gas state use Kp)
cC + dDaA + bB[A]a[B]b
[C]c[D]dKc =
(a) 2 BrCl (soln) →
Br2 (soln) + Cl2 (soln)
(b) 2 N2 (g) + 2 O2 (g) →
4 NO (g) (give Kp)
(c) BaSO4 (s) →
Ba 2+ (aq) + SO42-(aq)
(d) Fe2O3 (s) + 3 CO(g) →
2 Fe (s) + 3 CO2(g)
(s) or (l)
leave out of
K expression
HW 14.1: The Equilibrium Constant Kc
Write the equilibrium constant expression for the
following reactions. (Kp = equilibrium constants with amounts are given by
pressure of each component)(if gas is in solution use Kc, if gas is in gas state use Kp)
cC + dDaA + bB
[A]a[B]b
[C]c[D]dKc =
(a) 2 BrCl (soln) →
Br2 (soln) + Cl2 (soln)
(b) 2 N2 (g) + 2 O2 (g) →
4 NO (g) (give Kp)
(c) BaSO4 (s) →
Ba 2+ (aq) + SO42-(aq)
(d) Fe2O3 (s) + 3 CO(g) →
2 Fe (s) + 3 CO2(g)
[BrCl]2
[Br2] [Cl2]Kc =
(PN2)2(PO2)
2
(PNO)4
[Ba2+ ] [SO42-]
(PCO)3
( PCO2)3
Kp =
Kc =
Kp =
(s) or (l)
leave out of
K expression
2/28 Friday
C section
The Equilibrium Constant Kc
= 4.63 × 10–3
0.0429
(0.0141)2
Experiment 5
[N2O4]
[NO2]2
Kc == 4.64 × 10–3
0.0337
(0.0125)2
Experiment 1
The Equilibrium Constant Kc
Kc =
The equilibrium constant and the equilibrium constant
expression are for the chemical equation as written.
N2(g) + 3 H2(g)(b) 2 NH3(g)[NH3]
2
[N2][H2]3
=
4 NH3(g)(c) 2 N2(g) + 6 H2(g)[N2]
2[H2]6
[NH3]4
= Kc´´
1
Kc
2 NH3(g)(a) N2(g) + 3 H2(g)[N2][H2]
3
[NH3]2
Kc =
´
Kc =2
cC + dDaA + bB[A]a[B]b
[C]c[D]dKc =
reversed rxn (a) - turn K upside down
doubled rxn (a) – square all terms
reverse rxn double rxn
End 2/28F A
section
HW 14.2: The Equilibrium Constant Kc
Give the (a) equilibrium constant expression and the (b)
numerical value of the equilibrium constant expression for the
following reactions given the reaction. (assume dissolved in solvent)
N2 (g) + O2 (g) →
2 NO (g) Kc = 7.5 x 10 -9
(1) 2 NO (g) →
N2 (g) + O2 (g)
(2) 2 N2 (g) + 2 O2 (g) →
4 NO (g)
[N2] [O2]
[NO]2
Kc =
HW 14.2: The Equilibrium Constant KcGive the (a) equilibrium constant expression and the
(b) numerical value of the equilibrium constant
expression for the following reactions given the reaction.
N2 (g) + O2 (g) →
2 NO (g) Kc = 7.5 x 10-9
(1) 2 NO (g) →
N2 (g) + O2 (g)
Kc = 1/ 7.5 x 10 -9 = 1.33 x 10 8
(2) 2 N2 (g) + 2 O2 (g) →
4 NO (g)
Kc = (7.5 x 10-9)2 = 5.6 x 10 -17
[N2] [O2]
[NO]2
[N2]2 [O2]
2
[NO]4
Kc =
Kc =
[N2] [O2]
[NO]2
Kc =
The Equilibrium Constant Kc (Kp)
2 NO2(g)N2O4(g)
Kp =
N2O4P
NO2P
2
P is the partial pressure of that component.
HW 14.3: The Equilibrium Constant Kc (Kp)
CO (g) + H2O(g) → CO2 (g) + H2 (g)
What is the value of Kp at 700 K if the partial pressure in an
equilibrium mixture at 700 K are 1.31 atm of CO, 10.0 atm of
H2O, 6.12 atm of CO2 and 20.3 atm of H2 ?
HW 14.3: The Equilibrium Constant Kc (Kp)
CO (g) + H2O(g) →
CO2 (g) + H2 (g)
What is the value of Kp at 700 K if the partial pressure in an
equilibrium mixture at 700 K are 1.31 atm of CO, 10.0 atm of
H2O, 6.12 atm of CO2 and 20.3 atm of H2 ?
PCO2 PH2 (6.12 atm)(20.3 atm) = 9.48
Kp = =
PCO PH2O (1.31 atm)(10.0 atm)
The Equilibrium Constant Kc
Kp = Kc(RT)Dn
K mol
L atmR is the gas constant, 0.08206 .
T is the absolute temperature (kelvin).
Dn = # moles of gaseous products
minus # moles of gaseous reactants.
The Equilibrium Constant Kc
Kp = Kc(RT)Dn
Dn = (c+d) – (a+b)K mol
L atmR is the gas constant, 0.08206 .
2NO (g) + O2(g) → 2NO2 (g)
If Kc = 6.9 x 105 at 227oC, what is the value of Kp ?
Kp = Kc (RT)D n
Dn = (2) – (2+1) = -1
Kp = (6.9 x 105) [(0.08206 L atm/mol K)(227+ 273.15)] -1
Kp = 16811.9 = 1.7 x 104
HW 14.4: The Equilibrium Constant Kc
Kp = Kc(RT)Dn
Dn = (c+d) – (a+b) K mol
L atmR is the gas constant, 0.08206 .
H2O (g) + CH4(g) → CO (g) + 3H2 (g)
If Kc = 3.8 x 10-3 at 1000 K, what is the value of Kp ?
HW 14.4: The Equilibrium Constant Kc
Kp = Kc(RT)Dn
Dn = (c+d) – (a+b) K mol
L atmR is the gas constant, 0.08206 .
H2O (g) + CH4(g) → CO (g) + 3H2 (g)
If Kc = 3.8 x 10-3 at 1000 K, what is the value of Kp ?
Kp = Kc (RT)D n
Dn = (1+3) – (1+1) = 2
Kp = (3.8 x 10- 3) [(0.08206 L atm/mol K)(1000 K)] 2
Kp = 25.588 = 26
End 3/2 Monday
A section
Heterogeneous Equilibria
CaO(s) + CO2(g)CaCO3(s)
LimeLimestone
(1)
(1)[CO2]= [CO2]
[CaCO3]
[CaO][CO2]=
Pure solids and pure liquids are not
included in equilibrium expression.
Kc =
Kc = [CO2] Kp = PCO2
Molarity Pressure
If dissolved. If just gas – no solvent.
Using the Equilibrium Constant
(at 500 K) Kc = 4.2 × 10–48
2 H2(g) + O2(g)2 H2O(g)
(at 500 K) Kc = 2.4 × 1047
2 H2O(g)2 H2(g) + O2(g)
Kc = 57.0
2 HI(g)H2(g) + I2(g)
(at 700 K)
Using the Equilibrium Constant
cC + dDaA + bB
Qc =[A]t
a[B]tb
[C]tc[D]t
d
Reaction quotient:
The reaction quotient, Qc, is defined in the same way
as the equilibrium constant, Kc, except that the
concentrations in Qc are not necessarily equilibrium
values. (value of mixed reaction before equilibrium)
Using the Equilibrium Constant
• If Qc = Kc no net reaction occurs.
(at equilibrium)
• If Qc < Kc net reaction goes from left to right
(reactants to products, → ).
• If Qc > Kc net reaction goes from right to left
(products to reactants, ).
End 3/3 Tuesday
C section
HW 14.5: Using the Equilibrium Constant
• If Qc = Kc no net reaction occurs. (at equilibrium)
• If Qc < Kc net reaction goes from left to right (reactants to products, → ).
• If Qc > Kc net reaction goes from right to left (products to reactants, ).
If the equilibrium constant Kc = 2.78 x 10-10 and you have a reaction
A + 2 B → 3 C + D and [A] = 1.78 x 10-4, [B] = 1.1 x 10-5 and
[C] = 1.23 x 10 -6, [D] = 7.8 x 10-8 does the reaction go forward to
produce more products or backwards to produce more reactant or
stay in equilibrium ?
HW 14.5: Using the Equilibrium Constant
• If Qc = Kc no net reaction occurs. (at equilibrium)
• If Qc < Kc net reaction goes from left to right (reactants to products, → ).
• If Qc > Kc net reaction goes from right to left (products to reactants, ).
If the equilibrium constant Kc = 2.78 x 10-10 and you have a reaction
A + 2 B → 3 C + D and [A] = 1.78 x 10-4, [B] = 1.1 x 10-5 and
[C] = 1.23 x 10 -6, [D] = 7.8 x 10-8 does the reaction go forward to
produce more products or backwards to produce more reactant or
stay in equilibrium ?
Kc = Q =
Q = 6.74 x 10-12 & 6.74 x 10-12 < 2.78 x 10-10
Reaction goes forward until equilibrium is reached.
[A] [B]2
[C]3[D]
[1.78 x 10-4] [1.1 x 10-5]2
[1.23 x 10 -6]3[7.8 x 10-8]
End 3/2/20
Monday
section A
Using the Equilibrium Constant – ICE
method [Initial, Change, Equilibrium]
(given K & initial concentrations,
calculate equilibrium concentrations)
At 700 K, 0.500 mol of HI is added to a 2.00 L container
and allowed to come to equilibrium. Calculate the
equilibrium concentrations of H2, I2, and HI . Kc is 57.0
at 700 K.
2 HI(g)H2(g) + I2(g)
Initial: Zero
of reactants
Using the Equilibrium Constant
2 HI(g)H2(g) + I2(g)
Kc =[H2][I2]
[HI]2
0 0 0.250
+x +x –2x
x x 0.250 – 2x
57.0 =x2
(0.250 – 2x)2
Substitute values into the equilibrium expression:
Set up a table: Initial, Change, Equilibrium
Initial
Change
Equilibrium
0.500 mol in 2
L is 0.250 M
Using the Equilibrium Constant
Determine the equilibrium concentrations:
x = 0.0262x2
(0.250 – 2x)2
57.0 =
Solve for “x”:
HI = 0.250 – 2x = 0.250 – 2(0.0262) = 0.198 M
H2 = x = 0.0262 M
I2 = x = 0.0262 M
7.55 = 0.250 – 2x 7.55 x = 0.250 – 2x
x
9.55 x = 0.250 x = 0.250 /9.55 = 0.0262
HW 14.6: Using the Equilibrium Constant
2 HI(g)H2(g) + I2(g)
Kc = ______________________________________
Solution x = 0.00156 Hl = ____________________
Initial
Change
Equilibrium
For H2 𝑔 + l2 𝑔 ⇌ 2HI 𝑔 , for initial concentrations of H2(0.00623 𝑀), l2(0.00414 𝑀) and HI (0.0224 𝑀). Calculate the concentrations at equilibrium given
𝐾c (54.3) assuming forward rxn.
(a) Set up the ICE table. (where x = change of H2)
(b) show the expression for Kc
(c) If your solution to Kc gives x = 0.00156, calculate the concentration of HI.
HW 14.6: Using the Equilibrium Constant
2 HI(g)H2(g) + I2(g)
0.00623 0.00414 0.0224
-x -x +2x
0.00623-x 0.00414-x 0.0224 + 2x
[0.0224+2x]2 = 54.3
Kc = [0.00623-x][0.00414-x]
Solution x = 0.00156 Hl = 𝟎. 𝟎𝟐𝟐𝟒 + 𝟐 𝟎. 𝟎𝟎𝟏𝟓𝟔 = 𝟎. 𝟎𝟐𝟓𝟓𝑴
Initial
Change
Equilibrium
For H2 𝑔 + l2 𝑔 ⇌ 2HI 𝑔 , for initial concentrations of H2(0.00623 𝑀), l2(0.00414 𝑀) and HI (0.0224 𝑀). Calculate the concentrations at equilibrium given
𝐾c (54.3) assuming forward rxn.
(a) Set up the ICE table. (where x = change of H2)
(b) show the expression for Kc
(c) If your solution to Kc gives x = 0.00156, calculate the concentration of HI.
Step 2: The equilibrium constant is
54.3 =0.0224 + 2x 2
0.00623 − 𝑥 0.00414 − 𝑥
54.3 2.58 × 10−5 − 0.0104𝑥 + 𝑥2 = 5.02 × 10−4 +0.0896𝑥 + 4𝑥2
1.40 x 10-3 – 0.56472x + 54.3 x2 = 5.02 x 10-4 + 0.089x + 4x2
54.3 = 5.02 x 10-4 + 0.0448𝑥 + 0.0448𝑥 + 4𝑥 2
(2.58 x 10-5 – 0.00623 𝑥 -0.00414 𝑥 + 𝑥 2 )
50. 3𝑥2 − 0.654𝑥 + 8.98 × 10−4 = 0
Using the Equilibrium Constant – showing the math
not responsible for on exam – too long for quiz/exam
For equation: 50. 3𝑥2 − 0.654𝑥 + 8.98 × 10−4 = 0
This is a quadratic equation of the form a𝑥2 + 𝑏𝑥 + 𝑐 = 0. The
solution for a quadratic equation (see Appendix 4) is
𝑥 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
Here we have a = 50.3, 𝑏 = −0.654, and 𝑐 = 8.98 × 10−4
𝑥 =0.654 ± −0.654 2 − 4 50.3 8.98 × 10−4
2 × 50.3
𝑥 = 0.0114 𝑀 or 𝑥 = 0.00156 𝑀
Using the Equilibrium Constant – showing the math
not responsible for on exam – too long for quiz/exam
The first solution is physically impossible because the amounts
of H2 and l2 reacted would be more than those originally
present. The second solution gives the correct answer. Note
that in solving quadratic equations of this type, one answer is
always physically impossible, so choosing a value for 𝑥 is easy.
At equilibrium, the concentrations are
H2 = 0.00623 − 0.00156 𝑀 = 𝟎. 𝟎𝟎𝟒𝟔𝟕𝑴
I2 = 0.00414 − 0.00156 𝑀 − 𝟎. 𝟎𝟎𝟐𝟓𝟖 𝑴
Hl = 0.0224 + 2 × 0.00156 𝑀 = 𝟎. 𝟎𝟐𝟓𝟓 𝑴
Le Châtelier’s Principle
Le Châtelier’s Principle: If a stress is applied to a
reaction mixture at equilibrium, net reaction occurs in the
direction that relieves the stress. (possible changes)
(equilibrium shifts to undo stress applied to system)
• concentration of reactants or products
• pressure and volume
only effect for rxn with gases, otherwise not much
effect of P & V for liquid & solids
• temperature can be changed. (can predict only
if given enthalpy) (-DH exo → lower T)
Altering an Equilibrium Mixture: Changes in
Concentration
• the concentration stress of an added reactant or
product is relieved by net reaction in the direction
that consumes the added substance.
• the concentration stress of a removed reactant or
product is relieved by net reaction in the direction
that replenishes the removed substance.
In general, when an equilibrium is disturbed by the
addition or removal of any reactant or product,
Le Châtelier’s principle predicts that
Altering an Equilibrium Mixture: Changes in
Concentration
2 NH3(g)N2(g) + 3 H2(g) at 700 K, Kc = 0.291
Since Qc < Kc, more reactants will be consumed and the
net reaction will be from left to right.
(1.50)(3.00)3
(1.98)2
= 0.0968 < KcQc =[N2][H2]
3
[NH3]2
=
An equilibrium mixture of 0.50 M N2, 3.00 M H2, and
1.98 M NH3 is disturbed by increasing the N2
concentration to 1.50 M.
Altering an Equilibrium Mixture: Changes in Concentration
FeNCS2+(aq)Fe3+(aq) + SCN−(aq)
RedPale yellow Colorless
Add Addsubtract
subtract
add reactant, rxn goes → subtract reactant, rxn goes
add product, rxn goes subtract product, rxn goes →
End 3/4/20
A section,
C section
End Quiz 6
End Test 3
Altering an Equilibrium Mixture: Changes in Pressure &
Volume
2 NH3(g)N2(g) + 3 H2(g)
4 moles gas reactant → 2 moles gas product (this example)
* increase P or decrease V rxn goes → (toward side w less moles gas)
*decrease P or increase V rxn goes (toward side w more moles gas)
Altering an Equilibrium Mixture: Changes in
Pressure and Volume
• an increase in P (reducing V) - rxn goes toward
side with less moles gas
• a decrease in P (increasing V) - rxn goes toward
side with more moles gas
In general, when an equilibrium is disturbed by a change
in volume that results in a corresponding change in
pressure, Le Châtelier’s principle predicts that
Altering an Equilibrium Mixture: Changes in Temperature
2 NH3(g)N2(g) + 3 H2(g)DH° = –2043 kJ means
+ heat (heat is product)
As the temperature increases, the equilibrium shifts
from products to reactants.
exothermic reaction (-DHo) rxn releases heat
higher T = add heat rxn goes (to more reactant)
Altering an Equilibrium Mixture: Changes in
Temperature
• the equilibrium constant (K = product /reactant) for an
exothermic reaction (negative DH°) decreases (get
more reactant) as the temperature increases.
• the equilibrium constant for an endothermic reaction
(positive DH°) increases (get more product) as the
temperature increases.
In general, when an equilibrium is disturbed by a change
in temperature, Le Châtelier’s principle predicts that
End 3/6 A, C
section
The Effect of a Catalyst on Equilibrium
catalysis – no effect on equilibrium
activation energy for catalyzed path is lower that for uncatalyzed –
same activation energy for both → (forward) and (backward) reactions
HW 14.7: Le Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, net
reaction occurs in the direction that relieves the stress. (equilibrium shifts to undo stress applied to system)
2 A(g) + B(g) → 3 C(g) + 2 D(g) DH = negative (circle one parenthesis under
which direction will equilibrium shift each letter)
hint: DH negative is exothermic = (+ heat, heat is product)
(a) add A rxn goes [(→) or () ]
(b) remove B rxn goes [(→) or () ]
(c) add C rxn goes [(→) or () ]
(d) remove C rxn goes [(→) or () ]
(e) higher T rxn goes [(→) or () ] end Test 3 (c) & below
(f) higher P rxn goes [(→) or () ]
(assume all reactants & products in reaction are gases)
(g) higher V rxn goes [(→) or () ]
(assume all reactant & products in reaction are gases)
HW 14.7: Le Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, net
reaction occurs in the direction that relieves the stress. (equilibrium shifts to undo stress applied to system)
2 A (g) + B(g) →3 C(g) + 2 D(g) DH = negative (circle one parenthesis under
which direction will equilibrium shift each letter)
hint:DH negative is exothermic = (+ heat, heat is product)
(a) add A rxn goes [(→) or () ]
(b) remove B rxn goes [(→) or () ]
(c) add C rxn goes [(→) or () ]
(d) remove C rxn goes [(→) or () ]
(e) higher T rxn goes [(→) or () ] end Test 3 (e) & below
(higher T = add heat)
(f) higher P rxn goes [(→) or () ] (assume all reactants & products in reaction are gases) (moves to fewer moles gas)
(g) higher V rxn goes [(→) or () ] (assume all reactant & products in reaction are gases) (moves to fewer moles gas)
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